Question

Two workers are sliding 390 kg crate across the floor. One worker pushes forward on the crate with a force of 450 N while the other pulls in the same direction with a force of 200 N using a rope connected to the crate. Both forces are horizontal, and the crate slides with a constant speed. What is the crate's coefficient of kinetic friction on the floor?

Answer 1

Answer:

The coefficient of kinetic friction $\mu= 0.16989$

Explanation:

From Newton's second law

$\sum\overset{\rightarrow}{F}=m\cdot\overset{\rightarrow}{a}$

If the velocity is constant, that means the summation of all forces must be equal to zero. Draw the free-body diagram to obtain the sums of forces in x and y. It must include the Friction Force, in the opposite direction of the displacement, the weight ($W=mg=390*9.81=3825.9N$), the Normal Force, which is the is the consequence of Newton's third law and the forces from the two workers.

The sum in y is:

$\sum F_{y}=F_{N}-3825.9=0$

Solving for the $F_{N}$:

$F_{N}= 3825.\,\allowbreak9N$

The sum in x is:

$\sum F_{x}=450+200-F_{f}=0$

Solving for the $F_{f}$:

$F_{f}=650.0N$

The formula of the magnitude of the Friction force is

$F_{f}=\mu F_{N}$

That means the coefficient of friction is:

$\mu=\frac{F_{f}}{F_{N}}=\frac{650.0}{3825.\,\allowbreak9}=\allowbreak0.16989$